# Given DeltaH_f^@ for Al_2O_3, how much aluminum is formed if 838*kJ of energy are expended? 2Al(s) + 3/2O_2(g) rarr Al_2O_3(s) DeltaH^@=-1676*kJ*mol^-1

$A {l}_{2} {O}_{3} \left(s\right) + 1676$ $k J \rightarrow 2 A l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) .$
You have the balanced stoichiometric equation. If you had produced 1 mol of aluminum, then by the reaction's stoichiometry you would know that you would require a 1/2 mol quantity of alumina as a reactant, and also $838$ $k J$ energy. So treat the energy required as the limiting reagent. Please repost your answer here.